Lời giải:
a.
$|3x+1|=5$
$\Leftrightarrow 3x+1=\pm 5$
$\Leftrightarrow x=\frac{4}{3}$ hoặc $x=-2$
b.
$2|2x-3|=\frac{2}{5}$
$\Leftrightarrow |2x-3|=\frac{1}{5}$
$\Leftrightarrow 2x-3=\pm \frac{1}{5}$
$\Leftrightarrow x=\frac{8}{5}$ hoặc $x=\frac{7}{5}$
c.
$|2-3x|=|5-2x|$
$\Leftrightarrow 2-3x=5-2x$ hoặc $2-3x=2x-5$
$\Leftrightarrow x=-3$ hoặc $x=1,4$
\(a,\left|3x+1\right|=5\)
\(\left|3x+1\right|=\left\{{}\begin{matrix}3x+1khix\ge-\dfrac{1}{3}\\-3x-1khix< -\dfrac{1}{3}\end{matrix}\right.\)
Với \(x\ge-\dfrac{1}{3}\Rightarrow3x+1=5\Rightarrow3x=4\Rightarrow x=\dfrac{4}{3}\left(tm\right)\)
Với \(x< -\dfrac{1}{3}\Rightarrow-3x-1=5\Rightarrow-3x=6\Rightarrow x=-2\left(tm\right)\)
Vậy \(S=\left\{-2;\dfrac{4}{3}\right\}\)
\(b,2\left|2x-3\right|=\dfrac{2}{5}\)
\(\Leftrightarrow\left|2x-3\right|=\dfrac{1}{5}\)
\(\left|2x-3\right|=\left\{{}\begin{matrix}2x-3khix\ge\dfrac{3}{2}\\-2x+3khix< \dfrac{3}{2}\end{matrix}\right.\)
Với \(x\ge\dfrac{3}{2}\Rightarrow2x-3=\dfrac{1}{5}\Rightarrow2x=\dfrac{16}{5}\Rightarrow x=\dfrac{8}{5}\left(tm\right)\)
Với \(x< \dfrac{3}{2}\Rightarrow-2x+3=\dfrac{1}{5}\Rightarrow-2x=-\dfrac{14}{5}\Rightarrow x=\dfrac{7}{5}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{8}{5};\dfrac{7}{5}\right\}\)
\(c,\left|2-3x\right|=\left|5-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2-3x=5-2x\\2-3x=-5+2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=3\\-5x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{7}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-3;\dfrac{7}{5}\right\}\)
